Step 1: Understanding the Question:
We need to find the intersection point of a line defined by two points and a given plane equation.
Step 2: Key Formula or Approach:
The equation of a line passing through \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is:
\[ \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} = t \]
Step 3: Detailed Explanation:
The direction ratios of the line are \( (2-1, 2-1, 2-1) = (1, 1, 1) \).
The equation of the line is:
\[ \frac{x-1}{1} = \frac{y-1}{1} = \frac{z-1}{1} = t \]
From this, any point on the line can be written as \( (1+t, 1+t, 1+t) \).
Substitute these coordinates into the plane equation \( x+y+z=9 \):
\[ (1+t) + (1+t) + (1+t) = 9 \]
\[ 3 + 3t = 9 \Rightarrow 3t = 6 \Rightarrow t = 2 \]
Substitute \( t = 2 \) back into the parametric coordinates:
\[ x = 1+2 = 3, y = 1+2 = 3, z = 1+2 = 3 \]
Step 4: Final Answer:
The point of intersection is \( (3,3,3) \).